---
# System prepended metadata

title: LeetCode 21.Merge Two Sorted Lists
tags: [Recursion, Linked List]

---

###### tags: `Linked List`, `Recursion`

# LeetCode 21.Merge Two Sorted Lists

You are given the heads of two sorted linked lists ```list1``` and ```list2```.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.  

![圖片名稱](https://assets.leetcode.com/uploads/2020/10/03/merge_ex1.jpg)  
  
  
>Example 1:
```
Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]
```
>Example 2:
```
Input: list1 = [], list2 = []
Output: []
```
>Example 3:
```
Input: list1 = [], list2 = [0]
Output: [0]
```
 

### Constraints:

- The number of nodes in both lists is in the range ```[0, 50]```.
- $-100 <= Node.val <= 100$
- Both ```list1``` and ```list2``` are sorted in non-decreasing order.
---
### Idea:
>
### Solution:

Python:
```python=
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        merge = ListNode()
        cursor = merge
        while list1 and list2:
            if list1.val <= list2.val:
                cursor.next = list1
                list1 = list1.next
            else:
                cursor.next = list2
                list2 = list2.next
            cursor=cursor.next
        if list1:
            cursor.next = list1
        else:
            cursor.next = list2
        return merge.next
```

C++
```cpp=
```